Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ e^(-y) cos(y) dy

Accepted Answer

Find the indefinite integral using integration by parts. We have the integral of e to the negative x, sin x dx. Now because this says integration by parts, we'll use the formula. This is the integral UDV equals UV minus integral VDU, of course, plus C at the end. Now, in this case, we first need to pick U and DV. We'll let you in this case equals sin of x. Trig functions should be first when we're picking you. DV then will be E to the negative X DX. Now we just need V, which is the antiderivative of dV. This derivative is e to the negative x. DU then it's just a derivative of sine, which is cosine X. We now have u, which is sin x, multiplied by V e to the negative x. Minus the integral of negative e to the negative x multiplied. By duu, which is cosine x dx. Let's go ahead and rewrite. We have negative sin xe to the negative X. Minus, which will now turn to a plus integral, E to the negative X, cosine XDX. Now, we have another integration by parts with our E to the negative X, cosine XDX. Now we let you then for our new integral. Equals cosine x. DV equals E to the negative X. V is once again E to the negative X. And DU is a derivative of a cosine, which is negative sin. We have negative sin Xe to the negative X. Plus Our new UV e to the negative X, cosine X. Minus the integral E to the negative X multiplied by DU negative sin X D X. We cannot simplify everything here. Negative sin xe to the negative x minus cosine xe to the negative x. And then we have all of our negatives here. We have 3 negatives that will give us minus integral e to the negative x. Sign x dx. Now, we noticed something here. Our original integral was the integral of e to the negative xin xDX. We're going to keep going in a circle if we keep using integration by parts, but we notice here we have an integral e to g x, sin x dx. We can do a little algebra trick here by setting our integral equals to the integral of E to the negative Xin XDX. Now we can add. This integral e to the g x sin x to both sides. Now, we end up getting. 2 multiplied by your integral. Of sin xe to the g x minus cosine xe to the g x. We can now rewrite this by dividing by 2 the integral. Of E to the negative Xin X, DX equals sin xe to the g x minus cosine xe to the negative x divided by 2. Finally, to get our answer, we will just factor out E to the negative X. We then get negative sin x minus cosine X, E to the negative X, all divided by 2. We can then do one more simplification by factoring out a negative from the sine and cosine. To give us e to the g x divided by 2, sin x plus cosine x. And don't forget the plus C. This is the answer to our problem. OK, I hope to help you solve it. Thank you for watching. Goodbye.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ e^(-y) cos(y) dy

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ e^(-y) cos(y) dy